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![](https://rs.olm.vn/images/avt/0.png?1311)
Mg + 2HCl -> MgCl2 + H2 (1)
Fe + 2HCl -> FeCl2 + H2 (2)
mCu=0,48(g)
mMg;Fe=4,28-0,48=3,8(g)
nH2=0,02(mol)
Đặt nMg=a
nFe=b
Ta có:
\(\left\{{}\begin{matrix}24a+56b=3,8\\a+b=0,02\end{matrix}\right.\)
=>a=
sao âm nhỉ
![](https://rs.olm.vn/images/avt/0.png?1311)
* Cu + H2SO4 -> CuSO4 + H2O +SO2
* NaCl + H2O -> NaOH + Cl2 + H2
2 Fe + 3Cl2 -> 2FeCl3
* 2FeCl3 -> Cl2 + 2FeCl2
2NaCl--------->2Na+Cl2
Fe+Cl2--->FeCl2
2FeCl2+Cl2---->2FeCl3
Cu+H2SO4đ------------>CuSO4+SO2+H2O
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
Mg(HCO3)2 + 2HCl \(\rightarrow\) MgCl2 + 2CO2 + 2H2O
Ag2O + 2HCl \(\rightarrow\) 2AgCl + H2O
AgNO3 + HCl \(\rightarrow\) AgCl + HNO3
Fe + 2HCl \(\rightarrow\) FeCl2 + H2O
MgO + 2HCl \(\rightarrow\) MgCl2 + H2O
Cu(OH)2 + 2HCl \(\rightarrow\) CuCl2 + 2H2O
Mg + 2HCl \(\rightarrow\) MgCl2 + H2O
K2S + 2HCl \(\rightarrow\) 2KCl + H2S
NaHCO3 + HCl \(\rightarrow\) NaCl + CO2 + H2O
CaSO3 + 2HCl \(\rightarrow\) CaCl2 + SO2 + H2O
Na2O + 2HCl \(\rightarrow\) 2NaCl + H2O
Fe3O4 + 8HCl \(\rightarrow\) FeCl2 + 2FeCl3 + 4H2O
Ca(HSO3)2 + 2HCl \(\rightarrow\) CaCl2 + 2SO2 + 2H2O
Al(OH)3 + 3HCl \(\rightarrow\) AlCl3 + 3H2O
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
CuO + 2HCl \(\rightarrow\) CuCl2 + H2O
Ba(OH)2 + 2HCl \(\rightarrow\) BaCl2 + 2H2O
Fe(NO3)2 + 2HCl \(\rightarrow\) FeCl2 + 2HNO3
MgCO3 + 2HCl \(\rightarrow\) MgCl2 + CO2 + H2O
b.
BaCl2 + H2SO4 \(\rightarrow\) BaSO4 + 2HCl
Mg(HCO3)2 + H2SO4 \(\rightarrow\) MgSO4 + CO2 + 2H2O
Ag2O + H2SO4 \(\rightarrow\) Ag2SO4 + H2O
2AgNO3 + H2SO4 \(\rightarrow\) Ag2SO4 + 2HNO3
Fe + H2SO4 \(\rightarrow\) FeSO4 + H2
MgO + H2SO4 \(\rightarrow\) MgSO4 + H2O
Cu(OH)2 + H2SO4 \(\rightarrow\) CuSO4 + 2H2O
Mg + H2SO4 \(\rightarrow\) MgSO4 + H2
K2S + H2SO4 \(\rightarrow\) K2SO4 + H2S
2NaHCO3 + H2SO4 \(\rightarrow\) Na2SO4 + 2CO2 + H2O
CaSO3 + H2SO4 \(\rightarrow\) CaSO4 + SO2 + H2O
Na2O + H2SO4 \(\rightarrow\) Na2SO4 + H2O
Fe3O4 + 4H2SO4 \(\rightarrow\) FeSO4 + Fe2(SO4)3 + 4H2O
Ca(HSO3)2 + H2SO4 \(\rightarrow\) CaSO4 + SO2 + 2H2O
CaCl2 + H2SO4 \(\rightarrow\) CaSO4 + 2HCl
2Al(OH)3 + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 6H2O
Zn + H2SO4 \(\rightarrow\) ZnSO4 + H2
CuO + H2SO4 \(\rightarrow\) CuSO4 + H2O
Ba(OH)2 + H2SO4 \(\rightarrow\)BaSO4 + H2O
Fe(NO3)2 + H2SO4 \(\rightarrow\) FeSO4 + 2HNO3
MgCO3 + H2SO4 \(\rightarrow\) MgSO4 + CO2 + H2O
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{CH_3COOC_2H_5}=\dfrac{11}{88}=0,125\left(mol\right)\)
PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H+H_2O\)
0,125<------------------------------0,125
b, => mCH3COOH = 0,125.60 = 7,5 (g)
\(c,H=\dfrac{7,5}{12}.100\%=62,5\%\)
a.b.\(n_{CH_3COOC_2H_5}=\dfrac{11}{88}=0,125mol\)
\(CH_3COOH+C_2H_5OH\rightarrow\left(t^o,H_2SO_4\left(đ\right)\right)CH_3COOC_2H_5+H_2O\)
0,125 0,125 ( mol )
\(m_{CH_3COOH}=0,125.60=7,5g\)
c.\(n_{CH_3COOH}=\dfrac{12}{60}=0,2mol\)
\(H=\dfrac{0,125}{0,2}.100\%=62,5\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đề bảo "pư đạt trạng thái cân bằng" chứ không phải "phản ứng hoàn toàn" nên cần thêm dữ kiện để tính câu b) á :v
![](https://rs.olm.vn/images/avt/0.png?1311)
a) KOH: P2O5; HCl
HCl: FeO; Fe; Cu(OH)2
AgNO3: Fe
1) 6KOH + P2O5 → 2K3PO4 + 3H2O
2) KOH + HCl → KCl + H2O
3) 2HCl + FeO → FeCl2 + H2O
4) 2HCl + Fe → FeCl2 + H2
5) 2HCl + Cu(OH)2 → CuCl2 + 2H2O
6) 2AgNO3 + Fe → Fe(NO3)2 + 2Ag
b) HCl: Fe; NaOH; Mg(OH)2; FeO
NaOH: CuSO4; SO3
CuSO4: Fe
1) 2HCl + Fe → FeCl2 + H2
2) HCl + NaOH → NaCl + H2O
3) 2HCl + Mg(OH)2 → MgCl2 + 2H2O
4) 2HCl + FeO → FeCl2 + H2O
5) 2NaOH + CuSO4 → Na2SO4 + Cu(OH)2↓
6) 2NaOH + SO3 → Na2SO4 + H2O
7) CuSO4 + Fe → FeSO4 + Cu